SOLUTION: solve using the addition method 3x+2y=1 5x+3y=3 can you show me how to work this problem out?

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Question 42778This question is from textbook Intermediate Algebra
: solve using the addition method
3x+2y=1
5x+3y=3
can you show me how to work this problem out? This question is from textbook Intermediate Algebra

Found 2 solutions by Nate, tutorcecilia:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
-9x+-+6y+=+-3
+
10x+%2B+6y+=+6
=
-9x+%2B+10x+-+6y+%2B+6y+=+-3+%2B+6
x+=+3
PLUG:
3x+%2B+2y+=+1
3%283%29+%2B+2y+=+1
9+%2B+2y+=+1
2y+=+-8
y+=+-4
(3,-4)
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C++-1.5x+%2B+0.5%2C+%28-5%2F3%29x+%2B+1%29+

Answer by tutorcecilia(2152) About Me  (Show Source):
You can put this solution on YOUR website!
This problem involves solving two linear equations at the same time (simultaneously). In earlier algebra, you only had to solve only one equataion in order to find the x and/or the y value.
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When solving two linear equations at the same time, if the lines are perpendicular, the answer will give the same x and y value. The same x and y value shows where the lines cross.
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There are several methods to solve simultaneously.
1. Addition method
2. Substitution method
3. The matrix method
4. Graphing on a graphing calculator
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In the additon method, you are trying to get rid of one of the variables and solving for the remaining variable. Once you have the remaining variable, plug it back into either equation and solve for the missing variable.
Looking at both equations, what do you have to multiply each equation by in order to cause one variable to be subtracted? Every problem will be different:
If you multiply the first line by (-5) and the second line by (3), the x terms will subtract out. You would have "added" a negative plus a positive to equal zero.
(-5)(3x+2y=1)
(3)(5x+3y=3)
_____________
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Distribute the new number throughout the entire equation.
-15x-10y=-5
+15x+9y=+9
___________
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"Add" the two equations:
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0x-1y=4
Simply:
-1y=4
y=-4
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Plug y = -4 back into either of the original equations and solve for x:
3x+2y=1
3x + 2(-4)=1
3x-8=1
3x-8+8=1+8
3x=9
3x/3=9/3
x=3.
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Take the second equation (or take the first equation) and plug in the values for x and y.
5x+3y=3
5(3)+3(-4)=3
15-12=3
3=3
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Try the first equation:
3x+2y=1
3(3)+2(-4)=1
9-8=1
1=1
This also checks out, so the coordinate points where these two lines cross is (3, -4)